攻防世界-re部分题解（五）

ctf 17 minutes read (About 2612 words)

后面的题目也越来越难了，比较难比较复杂的题目会单拉出来写，这里就记录一些比较简单的题目。

easy_Maze

看名字就知道是个迷宫题，IDA打开看看

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int __cdecl main(int argc, const char **argv, const char **envp)
{
  __int64 v3; // rax
  int v5[7]; // [rsp+0h] [rbp-270h]
  int v6; // [rsp+C0h] [rbp-1B0h]
  int v7[7]; // [rsp+D0h] [rbp-1A0h]
  int v8; // [rsp+190h] [rbp-E0h]
  int v9[7]; // [rsp+1A0h] [rbp-D0h]
  int v10; // [rsp+1BCh] [rbp-B4h]
  int v11; // [rsp+1C0h] [rbp-B0h]
  int v12; // [rsp+1C4h] [rbp-ACh]
  int v13; // [rsp+1C8h] [rbp-A8h]
  int v14; // [rsp+1CCh] [rbp-A4h]
  int v15; // [rsp+1D0h] [rbp-A0h]
  int v16; // [rsp+1D4h] [rbp-9Ch]
  int v17; // [rsp+1D8h] [rbp-98h]
  int v18; // [rsp+1DCh] [rbp-94h]
  int v19; // [rsp+1E0h] [rbp-90h]
  int v20; // [rsp+1E4h] [rbp-8Ch]
  int v21; // [rsp+1E8h] [rbp-88h]
  int v22; // [rsp+1ECh] [rbp-84h]
  int v23; // [rsp+1F0h] [rbp-80h]
  int v24; // [rsp+1F4h] [rbp-7Ch]
  int v25; // [rsp+1F8h] [rbp-78h]
  int v26; // [rsp+1FCh] [rbp-74h]
  int v27; // [rsp+200h] [rbp-70h]
  int v28; // [rsp+204h] [rbp-6Ch]
  int v29; // [rsp+208h] [rbp-68h]
  int v30; // [rsp+20Ch] [rbp-64h]
  int v31; // [rsp+210h] [rbp-60h]
  int v32; // [rsp+214h] [rbp-5Ch]
  int v33; // [rsp+218h] [rbp-58h]
  int v34; // [rsp+21Ch] [rbp-54h]
  int v35; // [rsp+220h] [rbp-50h]
  int v36; // [rsp+224h] [rbp-4Ch]
  int v37; // [rsp+228h] [rbp-48h]
  int v38; // [rsp+22Ch] [rbp-44h]
  int v39; // [rsp+230h] [rbp-40h]
  int v40; // [rsp+234h] [rbp-3Ch]
  int v41; // [rsp+238h] [rbp-38h]
  int v42; // [rsp+23Ch] [rbp-34h]
  int v43; // [rsp+240h] [rbp-30h]
  int v44; // [rsp+244h] [rbp-2Ch]
  int v45; // [rsp+248h] [rbp-28h]
  int v46; // [rsp+24Ch] [rbp-24h]
  int v47; // [rsp+250h] [rbp-20h]
  int v48; // [rsp+254h] [rbp-1Ch]
  int v49; // [rsp+258h] [rbp-18h]
  int v50; // [rsp+25Ch] [rbp-14h]
  int v51; // [rsp+260h] [rbp-10h]

  v9[0] = 1;
  v9[1] = 1;
  v9[2] = -1;
  v9[3] = 1;
  v9[4] = -1;
  v9[5] = 1;
  v9[6] = -1;
  v10 = 0;
  v11 = 0;
  v12 = 0;
  v13 = 0;
  v14 = 1;
  v15 = -1;
  v16 = 0;
  v17 = 0;
  v18 = 1;
  v19 = 0;
  v20 = 0;
  v21 = 1;
  v22 = 0;
  v23 = -1;
  v24 = -1;
  v25 = 0;
  v26 = 1;
  v27 = 0;
  v28 = 1;
  v29 = -1;
  v30 = 0;
  v31 = -1;
  v32 = 0;
  v33 = 0;
  v34 = 0;
  v35 = 0;
  v36 = 0;
  v37 = 1;
  v38 = -1;
  v39 = -1;
  v40 = 1;
  v41 = -1;
  v42 = 0;
  v43 = -1;
  v44 = 2;
  v45 = 1;
  v46 = -1;
  v47 = 0;
  v48 = 0;
  v49 = -1;
  v50 = 1;
  v51 = 0;
  memset(v7, 0, 0xC0uLL);
  v8 = 0;
  memset(v5, 0, 0xC0uLL);
  v6 = 0;
  Step_0((int (*)[7])v9, 7, (int (*)[7])v7);
  Step_1((int (*)[7])v7, 7, (int (*)[7])v5);
  v3 = std::operator<<<std::char_traits<char>>(&_bss_start, "Please help me out!");
  std::ostream::operator<<(v3, &std::endl<char,std::char_traits<char>>);
  Step_2((int (*)[7])v5, 7);
  system("pause");
  return 0;
}

首先新建了三个地图，初始化了其中一个，然后利用下面两步分别初始化另外两个图

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Step_0((int (*)[7])v9, 7, (int (*)[7])v7);
Step_1((int (*)[7])v7, 7, (int (*)[7])v5);

具体的操作在Step_2((int (*)[7])v5, 7)里，可以看到只用了最后的一个图，可以直接动态调试dump出地图来，然后看一下具体操作

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__int64 __fastcall Step_2(int (*a1)[7])
{
  int v1; // eax
  __int64 v2; // rax
  __int64 v3; // rax
  __int64 result; // rax
  __int64 v5; // rax
  char v6[35]; // [rsp+10h] [rbp-30h]
  char v7; // [rsp+33h] [rbp-Dh]
  int v8; // [rsp+34h] [rbp-Ch]
  int v9; // [rsp+38h] [rbp-8h]
  int v10; // [rsp+3Ch] [rbp-4h]

  v10 = 0;
  v9 = 0;
  v8 = 0;
  while ( v8 <= 29 && (*a1)[7 * v10 + v9] == 1 )
  {
    std::operator>><char,std::char_traits<char>>(&std::cin, &v7);
    v1 = v8++;
    v6[v1] = v7;
    if ( v7 == 100 )
    {
      ++v9;
    }
    else if ( v7 > 100 )
    {
      if ( v7 == 115 )
      {
        ++v10;
      }
      else
      {
        if ( v7 != 119 )
          goto LABEL_14;
        --v10;
      }
    }
    else if ( v7 == 97 )
    {
      --v9;
    }
    else
    {
LABEL_14:
      v2 = std::operator<<<std::char_traits<char>>(&_bss_start, "include illegal words.");
      std::ostream::operator<<(v2, &std::endl<char,std::char_traits<char>>);
    }
  }
  if ( v10 != 6 || v9 != 6 )
  {
    v5 = std::operator<<<std::char_traits<char>>(&_bss_start, "Oh no!,Please try again~~");
    std::ostream::operator<<(v5, &std::endl<char,std::char_traits<char>>);
    result = 0LL;
  }
  else
  {
    v3 = std::operator<<<std::char_traits<char>>(&_bss_start, "Congratulations!");
    std::ostream::operator<<(v3, &std::endl<char,std::char_traits<char>>);
    output(v6, v8);
    result = 1LL;
  }
  return result;
}

难度不是很大，经典的wasd，只能经过1的位置，走到最后就可以了，整个路径的操作就是flag，格式为UNCTF{}

dump出的地图为

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1 0 0 1 1 1 1
1 0 1 1 0 0 1
1 1 1 0 1 1 1
0 0 0 1 1 0 0
1 1 1 1 0 0 0
1 0 0 0 1 1 1
1 1 1 1 1 0 1

走出来就可以了

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ssddwdwdddssaasasaaassddddwdds

验证一下

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root@rycbar17th:~# ./easy_Maze 
Please help me out!
ssddwdwdddssaasasaaassddddwdds
Congratulations!
Thanks! Give you a flag: UNCTF{ssddwdwdddssaasasaaassddddwdds}
sh: 1: pause: not found

得到flag

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UNCTF{ssddwdwdddssaasasaaassddddwdds}

ReverseMe-120

将输入的字符串进行base64加密然后异或，和明文字符串进行对比

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if ( v18 && *v19 >= v13 )
  {
    v21 = 3;
    v14 = 0;
    for ( lengtha = 0; v5; --v5 )
    {
      v15 = *v7;
      if ( *v7 != 13 && v15 != 10 && v15 != 32 )
      {
        v16 = byte_414E40[v15];
        v21 -= v16 == 64;
        v14 = v16 & 0x3F | (v14 << 6);
        if ( ++lengtha == 4 )
        {
          lengtha = 0;
          if ( v21 )
            *v12++ = BYTE2(v14);
          if ( v21 > 1 )
            *v12++ = BYTE1(v14);
          if ( v21 > 2 )
            *v12++ = v14;
        }
      }
      ++v7;
    }

有很多的无用代码，这里是关键代码，可以看到每四个字节进行了拼接，输出为三个字节，很显然是base64。

Replace

湖湘杯2018的题目，当时还参加过，那时候连这道题都毫无头绪(真的菜

这就是个简单的换表题，先upx脱个壳

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int __cdecl main(int argc, const char **argv, const char **envp)
{
  int v3; // kr00_4
  char Buf; // [esp+4h] [ebp-2Ch]
  char Dst; // [esp+5h] [ebp-2Bh]

  Buf = 0;
  memset(&Dst, 0, 0x27u);
  printf("Welcome The System\nPlease Input Key:");
  gets_s(&Buf, 0x28u);
  v3 = strlen(&Buf);
  if ( (unsigned int)(v3 - 35) <= 2 )
  {
    if ( sub_401090((int)&Buf, v3) == 1 )
      printf("Well Done!\n");
    else
      printf("Your Wrong!\n");
  }
  return 0;
}

signed int __fastcall sub_401090(int input, int length)
{
  int _input; // ebx
  int index; // edx
  char tmp_char; // al
  int v6; // esi
  int v7; // edi
  char v8; // al
  int v9; // eax
  char v10; // cl
  int v11; // eax
  int v12; // ecx

  _input = input;
  if ( length != 35 )
    return -1;
  index = 0;
  while ( 1 )
  {
    tmp_char = *(_BYTE *)(index + _input);
    v6 = (tmp_char >> 4) % 16;
    v7 = (16 * tmp_char >> 4) % 16;
    v8 = byte_402150[2 * index];
    if ( v8 < '0' || v8 > '9' )
      v9 = v8 - 'W';
    else
      v9 = v8 - '0';
    v10 = byte_402151[2 * index];
    v11 = 16 * v9;
    if ( v10 < '0' || v10 > '9' )
      v12 = v10 - 'W';
    else
      v12 = v10 - '0';
    if ( (unsigned __int8)byte_4021A0[16 * v6 + v7] != ((v11 + v12) ^ 0x19) )
      break;
    if ( ++index >= 35 )
      return 1;
  }
  return -1;
}

处理了半天，高四位低四位的处理来处理去，最后还要合在一起……

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target = [
    0x63, 0x7C, 0x77, 0x7B, 0xF2, 0x6B, 0x6F, 0xC5, 0x30, 0x01,
    0x67, 0x2B, 0xFE, 0xD7, 0xAB, 0x76, 0xCA, 0x82, 0xC9, 0x7D,
    0xFA, 0x59, 0x47, 0xF0, 0xAD, 0xD4, 0xA2, 0xAF, 0x9C, 0xA4,
    0x72, 0xC0, 0xB7, 0xFD, 0x93, 0x26, 0x36, 0x3F, 0xF7, 0xCC,
    0x34, 0xA5, 0xE5, 0xF1, 0x71, 0xD8, 0x31, 0x15, 0x04, 0xC7,
    0x23, 0xC3, 0x18, 0x96, 0x05, 0x9A, 0x07, 0x12, 0x80, 0xE2,
    0xEB, 0x27, 0xB2, 0x75, 0x09, 0x83, 0x2C, 0x1A, 0x1B, 0x6E,
    0x5A, 0xA0, 0x52, 0x3B, 0xD6, 0xB3, 0x29, 0xE3, 0x2F, 0x84,
    0x53, 0xD1, 0x00, 0xED, 0x20, 0xFC, 0xB1, 0x5B, 0x6A, 0xCB,
    0xBE, 0x39, 0x4A, 0x4C, 0x58, 0xCF, 0xD0, 0xEF, 0xAA, 0xFB,
    0x43, 0x4D, 0x33, 0x85, 0x45, 0xF9, 0x02, 0x7F, 0x50, 0x3C,
    0x9F, 0xA8, 0x51, 0xA3, 0x40, 0x8F, 0x92, 0x9D, 0x38, 0xF5,
    0xBC, 0xB6, 0xDA, 0x21, 0x10, 0xFF, 0xF3, 0xD2, 0xCD, 0x0C,
    0x13, 0xEC, 0x5F, 0x97, 0x44, 0x17, 0xC4, 0xA7, 0x7E, 0x3D,
    0x64, 0x5D, 0x19, 0x73, 0x60, 0x81, 0x4F, 0xDC, 0x22, 0x2A,
    0x90, 0x88, 0x46, 0xEE, 0xB8, 0x14, 0xDE, 0x5E, 0x0B, 0xDB,
    0xE0, 0x32, 0x3A, 0x0A, 0x49, 0x06, 0x24, 0x5C, 0xC2, 0xD3,
    0xAC, 0x62, 0x91, 0x95, 0xE4, 0x79, 0xE7, 0xC8, 0x37, 0x6D,
    0x8D, 0xD5, 0x4E, 0xA9, 0x6C, 0x56, 0xF4, 0xEA, 0x65, 0x7A,
    0xAE, 0x08, 0xBA, 0x78, 0x25, 0x2E, 0x1C, 0xA6, 0xB4, 0xC6,
    0xE8, 0xDD, 0x74, 0x1F, 0x4B, 0xBD, 0x8B, 0x8A, 0x70, 0x3E,
    0xB5, 0x66, 0x48, 0x03, 0xF6, 0x0E, 0x61, 0x35, 0x57, 0xB9,
    0x86, 0xC1, 0x1D, 0x9E, 0xE1, 0xF8, 0x98, 0x11, 0x69, 0xD9,
    0x8E, 0x94, 0x9B, 0x1E, 0x87, 0xE9, 0xCE, 0x55, 0x28, 0xDF,
    0x8C, 0xA1, 0x89, 0x0D, 0xBF, 0xE6, 0x42, 0x68, 0x41, 0x99,
    0x2D, 0x0F, 0xB0, 0x54, 0xBB, 0x16
]
var = [0x2a, 0x49, 0xf6, 0x9c, 0x38, 0x39, 0x5c, 0xde, 0x96, 0xd6, 0xde, 0x96, 0xd6, 0xf4, 0xe0, 0x25, 0x48, 0x49,
       0x54, 0xd6, 0x19, 0x54, 0x48, 0xde, 0xf6, 0xe2, 0xda, 0xd6, 0x77, 0x86, 0xe2, 0x1d, 0x5a, 0xda, 0xe6]
flag = ''
for i in var:
    tmp = target.index(i^0x19)
    flag += chr(tmp)
print(flag)

输出flag

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flag{Th1s_1s_Simple_Rep1ac3_Enc0d3}

babyRE

又是一道SMC，不知道为什么最近遇到的还是挺多的

没有任何加密，IDA打开

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int __cdecl main(int argc, const char **argv, const char **envp)
{
  char s; // [rsp+0h] [rbp-20h]
  int v5; // [rsp+18h] [rbp-8h]
  int i; // [rsp+1Ch] [rbp-4h]

  for ( i = 0; i <= 181; ++i )
  {
    envp = (const char **)(*((unsigned __int8 *)judge + i) ^ 0xCu);
    *((_BYTE *)judge + i) ^= 0xCu;
  }
  printf("Please input flag:", argv, envp);
  __isoc99_scanf("%20s", &s);
  v5 = strlen(&s);
  if ( v5 == 14 && (unsigned int)judge((__int64)&s) )
    puts("Right!");
  else
    puts("Wrong!");
  return 0;
}

经典的输入然后验证，不过judge函数是打不开的，看前面的操作，典型的SMC，写个IDC脚本跑一下

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#include <idc.idc>

static main() {
    auto judge, addr,tmp;
    judge = 0x600B00;
    auto i;
    for ( i = 0; i <= 181; ++i )
    {
        addr=judge+i;
        tmp=Byte(addr);
        tmp=tmp^ 0xC;
        PatchByte(addr,tmp);
    }

    Message("Down!");
}

然后改一下judge函数的结束位置，F5分析，逻辑很好懂

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signed __int64 __fastcall judge(__int64 a1)
{
  __int64 v2; // [rsp-8h] [rbp-8h]

  *(&v2 - 5) = a1;
  *((_BYTE *)&v2 - 32) = 0x66;
  *((_BYTE *)&v2 - 31) = 0x6D;
  *((_BYTE *)&v2 - 30) = 0x63;
  *((_BYTE *)&v2 - 29) = 0x64;
  *((_BYTE *)&v2 - 28) = 0x7F;
  *((_BYTE *)&v2 - 27) = 0x6B;
  *((_BYTE *)&v2 - 26) = 0x37;
  *((_BYTE *)&v2 - 25) = 0x64;
  *((_BYTE *)&v2 - 24) = 0x3B;
  *((_BYTE *)&v2 - 23) = 0x56;
  *((_BYTE *)&v2 - 22) = 0x60;
  *((_BYTE *)&v2 - 21) = 0x3B;
  *((_BYTE *)&v2 - 20) = 0x6E;
  *((_BYTE *)&v2 - 19) = 0x70;
  for ( *((_DWORD *)&v2 - 1) = 0; *((_DWORD *)&v2 - 1) <= 13; ++*((_DWORD *)&v2 - 1) )
    *(_BYTE *)(*((signed int *)&v2 - 1) + *(&v2 - 5)) ^= *((_DWORD *)&v2 - 1);
  for ( *((_DWORD *)&v2 - 1) = 0; *((_DWORD *)&v2 - 1) <= 13; ++*((_DWORD *)&v2 - 1) )
  {
    if ( *(_BYTE *)(*((signed int *)&v2 - 1) + *(&v2 - 5)) != *((_BYTE *)&v2 + *((signed int *)&v2 - 1) - 32) )
      return 0LL;
  }
  return 1LL;
}

这里其实有三个变量，&v2-1的位置是一个整型，实际上是索引值，&v2-5是我们输入的字符串，&v2-19是最终的目标，只有一个异或操作，很简单就可以跑出来结果

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target = [0x66, 0x6D, 0x63, 0x64, 0x7F, 0x6B, 0x37, 0x64, 0x3B, 0x56, 0x60, 0x3B, 0x6E, 0x70]
flag = ''
for i in range(14):
    flag += chr(target[i] ^ i)
print(flag)

输出flag

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flag{n1c3_j0b}

What-does-this-button-do

不知道为什么这题放在了这个位置，简单得不能再简单的一个mobile

jadx打开，很快就可以找到关键代码

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public void onCreate(Bundle savedInstanceState) {
       super.onCreate(savedInstanceState);
       setContentView(R.layout.activity_flag);
       String flag = "";
       int[] d = {102, 108, 97, 103, 123, 119, 52, 110, 110, 52, 95, 106, 52, 114, 95, 109, 121, 95, 100, 51, 120, 125};
       for (int i = 0; i < 22; i++) {
           flag = flag.concat(String.valueOf((char) d[i]));
       }
       ((TextView) findViewById(R.id.flagText)).setText(flag);
   }

然后直接输出把flag输出就可以了

不过这题要去掉外面的flag{}提交

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w4nn4_j4r_my_d3x

76号

单字检测，pintools爆破，得到flag

直接滚轮子

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flag{09vdf7wefijbk}

easy_go

这题根据名字也知道是go逆向，IDA打开是乱七八糟的函数，也没有任何的symbol，用go重命名的插件给函数重命名，发现一个base64的解码函数和奇怪的字符串，字母表不是很清楚，动态调试一下

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[-------------------------------------code-------------------------------------]
   0x4952dd:	mov    QWORD PTR [rsp],rax
   0x4952e1:	mov    QWORD PTR [rsp+0x8],rcx
   0x4952e6:	mov    QWORD PTR [rsp+0x10],rdx
=> 0x4952eb:	call   0x47e620
   0x4952f0:	mov    rax,QWORD PTR [rsp+0x38]
   0x4952f5:	mov    rcx,QWORD PTR [rsp+0x30]
   0x4952fa:	mov    rdx,QWORD PTR [rsp+0x18]
   0x4952ff:	mov    rbx,QWORD PTR [rsp+0x20]
No argument
[------------------------------------stack-------------------------------------]
0000| 0xc000070e90 --> 0xc000074580 ("6789_-abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ012345", '\377' <repeats 45 times>, "\005\377\377:;<=>?")
0008| 0xc000070e98 --> 0xc000088040 ("tGRBtXMZgD6ZhalBtCUTgWgZfnkTgqoNsnAVsmUYsGtCt9pEtDEYsql3")
0016| 0xc000070ea0 --> 0x38 ('8')
0024| 0xc000070ea8 --> 0x40 ('@')
0032| 0xc000070eb0 --> 0xc000088040 ("tGRBtXMZgD6ZhalBtCUTgWgZfnkTgqoNsnAVsmUYsGtCt9pEtDEYsql3")
0040| 0xc000070eb8 --> 0x38 ('8')
0048| 0xc000070ec0 --> 0x1 
0056| 0xc000070ec8 --> 0x1 
[------------------------------------------------------------------------------]
Legend: code, data, rodata, value

Thread 1 "easyGo" hit Breakpoint 3, 0x00000000004952eb in ?? ()

找到字母表和密文，解密得到flag

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flag{92094daf-33c9-431e-a85a-8bfbd5df98ad}

easyCpp

这题有很多c++标准库的调用，整体来说不是很难，可以通过调试发现每一部分的作用，也不是很难

re5-packed-movement

极其无聊的一道题，硬生生用mov实现了所有操作，要求输入flag对比一下，直接objdump就能再汇编代码里找到逐字符将flag移入到某个地址，或者因为是逐字符验证，直接pintools爆破也可以

the_maya_society

大致意思就是对当前时间进行一个md5加密，然后和一个字符串连接成目标网址，如果这个目标是对的后面的通信就会得到正确的flag……

时间应该是2012-12-21

2ex1

mips，并且strip过，什么信息都没有，打开输出发现很像base64，但应该换过表，ghidra打开找找可疑字符串，发现奇怪的字符串

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@,.1fgvw#`/2ehux$~"3dity%_;4cjsz^+{5bkrA&=}6alqB*-[70mpC()]89noD

试一试就得到了flag

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flag{change53233}

对这个字符串表进行交叉引用查询到话也可以找到进行base64编码的函数

#ctf #re #攻防世界
python_opcode IDA常用宏定义

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